Solveeit Logo
Login

Question

Question: How do you verify the identity \(\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}...

How do you verify the identity sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y=tan⁡x+tan⁡y1−tan⁡xtan⁡y\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} ?

Explanation

Solution

To solve this question, we need to use the basic relations of the trigonometric functions. We will mainly use the relation between sine, cosine and tangent function. This relation is that the ratio of sine and cosine function is the tangent function.

Complete step by step answer:
L.H.S=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡yL.H.S = \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}}
First, we will divide both numerator and denominator by cos⁡xcos⁡y\cos x\cos y.
⇒L.H.S=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow L.H.S = \dfrac{{\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y}}}}{{\dfrac{{\cos x\cos y - \sin x\sin y}}{{\cos x\cos y}}}}
Now, we will separate the terms in numerator and denominator.
⇒L.H.S=sin⁡xcos⁡ycos⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow L.H.S = \dfrac{{\dfrac{{\sin x\cos y}}{{\cos x\cos y}} + \dfrac{{\cos x\sin y}}{{\cos x\cos y}}}}{{\dfrac{{\cos x\cos y}}{{\cos x\cos y}} - \dfrac{{\sin x\sin y}}{{\cos x\cos y}}}}
We will now cancel out similar terms from numerator and denominator.
⇒L.H.S=sin⁡xcos⁡x+sin⁡ycos⁡y1−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow L.H.S = \dfrac{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin y}}{{\cos y}}}}{{1 - \dfrac{{\sin x\sin y}}{{\cos x\cos y}}}}
We know that the ratio of sine and cosine function is the tangent function.
Therefore, sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x and sin⁡ycos⁡y=tan⁡y\dfrac{{\sin y}}{{\cos y}} = \tan y.
We will put these values in the L.H.S.
⇒L.H.S=tan⁡x+tan⁡y1−tan⁡xtan⁡y\Rightarrow L.H.S = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}
This is our R.H.S.
∴L.H.S=R.H.S.\therefore L.H.S = R.H.S.

Hence, it is proved that sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y=tan⁡x+tan⁡y1−tan⁡xtan⁡y\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}.

Note: Here, we have started with the left hand side and then proved the given identity. But, we can also start with the right hand side and reach to the left hand side to prove the given trigonometric identity.
⇒R.H.S=tan⁡x+tan⁡y1−tan⁡xtan⁡y\Rightarrow R.H.S = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}
We will first use the definition of tangent function that it is the ratio of sine and cosine functions.Therefore,
sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x and tan⁡y=sin⁡ycos⁡y\tan y = \dfrac{{\sin y}}{{\cos y}}
We will put these values in the R.H.S.
⇒R.H.S=sin⁡xcos⁡x+sin⁡ycos⁡y1−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow R.H.S = \dfrac{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin y}}{{\cos y}}}}{{1 - \dfrac{{\sin x\sin y}}{{\cos x\cos y}}}}
Now we will take LCM in both numerator and denominator. The LCM for both is cos⁡xcos⁡y\cos x\cos y.
⇒R.H.S=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow R.H.S = \dfrac{{\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y}}}}{{\dfrac{{\cos x\cos y - \sin x\sin y}}{{\cos x\cos y}}}}
We can also write this term by converting the division into multiplication.
⇒R.H.S=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y×cos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡y\Rightarrow R.H.S = \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y}} \times \dfrac{{\cos x\cos y}}{{\cos x\cos y - \sin x\sin y}}
We know that the common terms in the numerator and the denominator will get cancelled out. Therefore, here cos⁡xcos⁡y\cos x\cos y will be cancelled out and we can write the right hand side as:
⇒R.H.S=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y\Rightarrow R.H.S = \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}}
This is our R.H.S.
⇒R.H.S=L.H.S\Rightarrow R.H.S = L.H.S
Hence, it is proved that sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y=tan⁡x+tan⁡y1−tan⁡xtan⁡y\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}.